\problemdata      
        {counting-5} % latex-friendly label for the prob.
        {Counting} % The topic(s) of the problem content.
        {F99 TU16} % Source (if known)
        {S02 CP8W; S99 TU9; F99 TU16} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem} We just demonstrated how to ``guess'' the fifth
card in a poker hand when a collaborator reveals the other four cards.
Describe a method for guessing two hidden cards in a hand of nine
cards, when your collaborator reveals the other seven cards.

\solution{
There must be $\ceil{9/4} = 3$ cards with the same suit, so our
collaborator chooses to hide two of them and then use the third one as the
first card to be revealed.  So this first revealed card fixes the suit of
the two hidden cards; it will also be used as the origin for the offset
position of the first hidden card.  This first hidden card will in turn be
used as the origin for the offset of the other hidden card.  There are six
cards to code the two offset positions.  These suffice to code two offsets
of size from one to six.  That is, our collaborator can choose one of the
$3! = 6$ orders in which to reveal the first three cards and thereby tell
us the offset position of the first hidden card.  Our collaborator can
then choose the order of the final three cards to describe the offset
position of the second hidden card from the first.  Note that the first
revealed card must be chosen so that both offsets are less $\leq 6$; since
the sum of the offsets between successive cars ordered in a cycle from Ace
to King is 13, it is not possible for more than one offset between
successive cards to exceed seven, so this is always possible.
}

% \problempart Show that there is no procedure for ``guessing'' a hidden card
% among four when your collaborator reveals three. \hint Compare the number
% of hands to the number of revealed three-card sequences.
% ~
% \solution{
% There are $\binom{52}{4}$ possible four-card hands.  There are $P(52,3)$
% three-card sequences that can ever be revealed.  Since
% \[
% \frac{\binom{52}{4}}{P(52,3)}=
% \frac{52\cdot 51\cdot 50 \cdot 49/4!}{52 \cdot 51 \cdot 50} = 49/24 > 1,
% \]
% there must, by pigeonholing, be two (actually three) hands for which the
% collaborator reveals the same three cards.  Therefore the magician cannot
% determine the hidden card, since these hands must have different hidden
% cards.}
\end{problem}
